According to the source below, the solubility of sulfanilamide in 95% ethyl alcohol at 78°C is 210 mg/mL. Since 0.1 g = 100 mg, we can set up a proportion:
(210 mg) / (1 mL) = (100 mg) / (x mL) Solving, x = 0.48 mL of 95% ethyl alcohol will be required.
I do not know previously the solubility of sulfanilamide in 95% ethyl alcohol. Let us accept the solubility you quoted here.
100/210 = 0.47619047619.. ≈ 0.48 (ml)
at 0C, the amount of sulfanilamide remains in the solution is: 14*(100/210) = 6.67 (mg), since you only have 0.48 ml solution.
The volume of the solution will change a little by cooling from 78C to 0C. You may also consider this volume change if you have data.
Calcium chloride contains only one calcium and 2 chloride atom thus it's ionic formula would be CaCl2.
Your question is incomplete. However, I found a similar problem fromanother website as shown in the attached picture.
To solve this problem, you must know that at STP, the volume for any gas is 22.4 L/mol. So,
Moles O₂: 156.8 mL * 1 L/1000 mL* 1 mol/22.4 L = 0.007 moles
Mass calcium: 0.007 mol O₂ * 2 mol Ca/1 mol O₂ * 40 g/mol Ca =
<em> 0.56 g Ca</em>
Answer : The current passing between the electrodes is, 
Explanation :
First we have to calculate the charge of sodium ion.

where,
q = charge of sodium ion
n = number of sodium ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the charge of chlorine ion.

where,
q' = charge of chlorine ion
n = number of chlorine ion = 
e = charge on electron = 
Now put all the given values in the above formula, we get:

Now we have to calculate the current passing between the electrodes.



Thus, the current passing between the electrodes is, 